One can prove that \(\mathcal {L}\)  is a novel (consensus) high-order Laplace matrix Let us introduce \(\mathcal {L}^{\mathit {sym}}=\mathbf {d}^{-1/2}\mathbf {L}^H\mathbf {d}^{-1/2}\) , where \(\mathbf {d}\)  is the diagonal matrix containing the \(d_i\) 's on the diagonal and \(\mathbf {L}^H\)  is the high-order (combinatorial) Laplace matrix defined in [1]}. Then \(\mathcal {L}^{\mathit {sym}}=D_{ij}/\sqrt{d_i d_j} -\delta _{ij}\)  from which it immediately follows that \(\mathcal {L}^{\mathit {sym}}\) 
is symmetric and nonpositive definite; indeed take any \(\mathbf {x}\in \mathbb {R}^N\setminus \lbrace 0\rbrace \) , \(N\)  standing for the dimension of the matrices,
then \((\mathbf {x},\mathcal {L}^{\mathit {sym}} \mathbf {x})=(\mathbf {d}^{-1/2} \mathbf {x},\mathbf {L}^H\mathbf {d}^{-1/2} \mathbf {x})\le 0\)  where the last inequality follows from the fact that \(\mathbf {L}^H\)  is nonpositive definite. Finally let us observe that \(\mathcal {L}=\mathbf {d}^{-1}\mathbf {L}^H =\mathbf {d}^{-1/2}\mathcal {L}^{\mathit {sym}}\mathbf {d}^{1/2}\) , hence, \(\mathcal {L}\)  is similar to \(\mathcal {L}^{\mathit {sym}}\)  and, thus they display the same non-positive spectrum. Moreover this implies also that \(-2\le \Lambda ^{(\alpha )}\le 0\) . , i.e. it is nonpositive definite, the largest eigenvalue is \(\Lambda ^{(1)}=0\)  and its is associated to the uniform eigenvector \(\phi ^{(1)}\sim (1,\dots ,1)^{\top }\) .
